How to find all the positive integers n of 4 digits such that all its digits are perfect squares and n is a multiple of 2, 3, 5 and 7? I was trying to use the divisibility rules of 2,3,5 and 7 but I becomes very tedious and couldn's solve the problem. I think there could be a faster way to solve it or to apply those rules. Please help me and thank you very much.
 A: Hint: The last digit has to be $0$ to be divisible by $2$ and $5$ simultaneously. Thus, The number you are seeking is $ABC0$ where $A,B,C\in\{0,1,4,9\}$ but $A \ne 0$.
Also, $A+B+C=3k$ for divisibility via $3$-Equation $(1)$
And $A+B+C =7 \lambda$-Equation$(2)$
Although I'd recommend checking divisibility rules for $7$ before taking equation $(2)$ into account.
or in short, it should be divisible by $21$.
I can see only one number here i.e. $4410$
Disclaimer: But there can be more. I haven't check the whole sample space yet.
A: If $n$ is divisible by $2,3,5$ and $7$, then $n$ is a multiple of 
$2 \cdot 3 \cdot 5 \cdot 7 \cdot = 210$.
That means that $m = \dfrac{1}{10}n = abc_{10}$ is a multiple of $21$ and 
$\{a,b,c\} \subset \{0,1,4,9\}$.
Let's first find all $m$ thst are multiples of $3$.
Note that $0, 9 \equiv 0 \pmod 3$ and $1,4 \equiv 1 \pmod 3$ So $m$ consists of all $1$'s and $4$'s or of all $0$'s and $9$'s. We get these possibilities.
\begin{array}{cccccccc}
111 & 114 & 141 & 144 & 411 & 414 & 441 & 444 \\
900 & 909 & 990 & 999
\end{array}
Out of these $12$ integers, only $441$ is also a multiple of $7$.
Hence the answer is $n = 4410$
